Topology and Euler Classes of Oriented 2-Plane Bundles in R^4

[lavinia]

I'd like some help understanding three things.

- What is the topology of the Grassmann manifold or oriented 2-planes in R^4?

- What is the Euler class of the canonical oriented 2 plane bundle over it?

- I think that the Euler class of the normal bundle of an oriented surface in R^4 is zero. Is the normal bundle always trivial?

 

[lavinia]

I'd like some help understanding three things.
- What is the topology of the Grassmann manifold or oriented 2-planes in R^4?

An oriented 2-plane can be represented by the wedge product of two orthonormal vectors.

Writing these vectors in orthonormal coordinates, this wedge product is a vector in 6 dimensions whose coordinates's sum of squares add to one and have one other linear relation.

This 4 manifold maps onto a Cartesian product of two 2-spheres using a simple mapping. This product of 2-spheres is just the diagonal product when R$$^{6}$$ is viewed as R$$^{3}$$ $$\times$$R$$^{3}$$

The mapping is linear from R$$^{6}$$ into itself and so is continuous from the 4 manifold onto its image. The mapping is also an isomorphism so it is a homeomorphism of the 4 manifold onto the product of two 2-spheres.

Details omitted since I am talking to myself.

 

[lavinia]

An oriented 2-plane can be represented by the wedge product of two orthonormal vectors.

Writing these vectors in orthonormal coordinates, this wedge product is a vector in 6 dimensions whose coordinates's sum of squares add to one and have one other linear relation.

This 4 manifold maps onto a Cartesian product of two 2-spheres using a simple mapping. This product of 2-spheres is just the diagonal product when R$$^{6}$$ is viewed as R$$^{3}$$ X R$$^{3}$$

The mapping is linear from R$$^{6}$$ into itself and so is continuous from the 4 manifold onto its image. The mapping is also an isomorphism so it is a homeomorphism of the 4 manifold onto the product of two 2-spheres.

Details omitted since I am talking to myself.

Maybe this works as another way to look at it.

Each element of SO(4) act on the 3 sphere. Because it acts by isometries, SO$$^{4}$$ acts on the tangent 2 sphere bundle to S$$^{3}$$. The derivative of this action acts on the tangent bundle to the tangent two sphere bundle and is also an isometry. Now look at the tangent circles to the tangent 2 spheres. SO(4) acts transitively and without fixed points on this circle bundle.

Therefore the tangent 2 sphere bundle is homeomorphic to SO$$^{4}$$ / SO$$^{1}$$ and this quotient is homeomorphic to S$$^{3}$$ X S$$^{2}$$ because the tangent bundle of the 3 sphere is trivial. But SO$$^{4}$$/SO$$^{1}$$ is just the Stiefel manifold of oriented 2 frames in R$$^{4}$$.

The quotient, SO$$^{4}$$ / SO$$^{1}$$ X SO$$^{1}$$ is the Grassman of oriented 2 planes and is isomorphic to a product of two 2 spheres. This last point need to be hardened.

 

[lavinia]

The quotient, SO$$^{4}$$ / SO$$^{1}$$ X SO$$^{1}$$ is the Grassman of oriented 2 planes and is isomorphic to a product of two 2 spheres. This last point need to be hardened.

Take the natural trivialization of the tangent bundle then it becomes clear that one is projecting along Hopf fibrations.

 

[lavinia]

I'd like some help understanding three things.
- What is the Euler class of the canonical oriented 2 plane bundle over it?

The Euler class is the sum of the two generators of the second cohomology of the product of two spheres. The proof is a little tricky.

 

[zhentil]

First, SO(1) is a point - are you thinking of U(1)?

For the Euler class of the normal bundle, you know that the first Chern class of the pullback of the tangent bundle of R^4 has to be trivial. This tells you that the first chern class (i.e. Euler class) is minus the Euler class of the surface.

To understand the isomorphism, I think the best analogy is that SO(4) clearly acts transitively on oriented 2-planes. The stabilizer is the stabilizer of the 2-plane crossed with the action on the orthogonal complement (i.e. S^1 x S^1).

 

[lavinia]

First, SO(1) is a point - are you thinking of U(1)?

For the Euler class of the normal bundle, you know that the first Chern class of the pullback of the tangent bundle of R^4 has to be trivial. This tells you that the first chern class (i.e. Euler class) is minus the Euler class of the surface.

To understand the isomorphism, I think the best analogy is that SO(4) clearly acts transitively on oriented 2-planes. The stabilizer is the stabilizer of the 2-plane crossed with the action on the orthogonal complement (i.e. S^1 x S^1).

Right. I wanted to use the Gysin sequence of the bundle of oriented 2-planes to caculate its Euler class. Once it is shown that the grassman is a product of two 2-spheres and the unit circle bundle is a product of a 2-sphere with a 3-sphere the Gysin sequence looks like

0 $$\rightarrow$$ H$$^{0}$$(Grassman of oriented 2-planes in R$$^{4})$$ $$\stackrel{}{\rightarrow}$$ H$$^{2}$$(Grassman) $$\rightarrow$$ H$$^{2}$$(Unit circle bundle) $$\rightarrow$$ 0

The second arrow is cup product with the Euler class so it can be computed by computing the kernel of the next map in the sequence. This is the problem of finding the homology image of a generator of the second homology of the unit circle bundle under the bundle projection map. calculation in coordinates shows that the image is the difference of the two homology generators so the Euler class is the sum of the cohomology generators.

It did not seems easy to show this from the action of SO(4) on the oriented 2-planes or otherwise put it was not obvious to me why SO(4)/U(1) x U(1) was a product of 2 two spheres. So I did it in coordinates instead. Perhaps you could show me how to see directly that SO(4)/U(1) x U(1) is a product of two 2-spheres.

And more generally how can one attack the Gysin sequence for other cases e.g. oriented 4 planes is 8-space? I would appreciate a general way to look at this.

Also, I was just reading in Milnor's book that the cohomology of the infinite Grassman of orented 2 planes is generated by the Euler class. So I guess one could do the same computation by looking at the inclusion map of the Grassman of oriented 2-planes in 4 space into the infinite Grassman. I haven't done this yet but it seems like a good exercise.

 

[lavinia]

For the Euler class of the normal bundle, you know that the first Chern class of the pullback of the tangent bundle of R^4 has to be trivial. This tells you that the first chern class (i.e. Euler class) is minus the Euler class of the surface.

I don't see your argument. It seems to me that the Euler class of the normal bundle must be zero. An oriented surface bounds in 4 space so the Thom class of the normal bundle must be null homologous.

It seems that your Whitney sum argument shows that the Euler characteristic of the surface must be even for otherwise the normal bundle would have to have a non zero second Stiefel-Whitney class which I believe must be zero.

Here is another argument or really the same argument.

In the exact sequence of the normal bundle,N,

H$$^{1}$$(N-0) $$\rightarrow$$ H$$^{2}$$(N,N-0) $$\rightarrow$$ H$$^{2}$$(N)

(Over the integers, this sequence is H$$^{1}$$(N-0) $$\rightarrow$$ Z $$\rightarrow$$ Z)

the first arrow must be the zero map if the Euler class is non-zero. But this means that the fiber circles are boundaries in N-0. But this can not happen since the surface is a boundary in R$$^{4}$$.

if these arguments are right, the first Chern class argument is not - and visa versa - so I am not sure what is going on at the moment.

 

[zhentil]

Yikes! What I said is rubbish. It depends on a holomorphic embedding, which we certainly don't have. What you said is correct - I can think of a transversality approach (which seems too complicated), but I'll think on it.

For the isomorphism between this and S^2 x S^2, let me know what you think of this. SO(4) acts transitively on S^2 x S^2 as follows. Pick a point (x,y) in S^2 x S^2 and take the oriented plane in R^3 perpendicular to it, for each. SO(3) acts transitively on each of these - let A in SO(3) act on the plane, and define A.x to be the unique point on S^2 such that the triple forms an oriented basis of R^3. Now examine the stabilizer of (North Pole, North Pole) - it's clear a copy of S^1 in each of the SO(3)'s. So we obtain (SO(3) x SO(3)) / (U(1) x U(1)) is diffeomorphic to S^2 x S^2. Since SO(4) is diffeomorphic to SO(3) x SO(3), we're done.

 

[lavinia]

It depends on a holomorphic embedding, which we certainly don't have. I can think of a transversality approach ...

So we obtain (SO(3) x SO(3)) / (U(1) x U(1)) is diffeomorphic to S^2 x S^2. Since SO(4) is diffeomorphic to SO(3) x SO(3), we're done.

In the first part you are saying that with a holomorphic embedding of the surface into R$$^{4}$$ one can pull back the tangent bundle of R$$^{4}$$ with its complex structure to get a trivial complex vector bundle over the surface. The tangent bundle to the surface is a complex subbundle and it's complement using a Hermitian metric - I suppose the metric is induced from R$$^{4}$$ - is another complex subbundle. Their Whitney sum is a trivial complex vector bundle.

- If the Chern class argument is wrong what can not happen? There is no holomorphic embedding of a surface into R$$^{4}$$ no matter what complex structure the surface and R$$^{4}$$ are given?

In the second part, I see that (SO(3) x SO(3)) / (U(1) x U(1)) is diffeomorphic to S^2 x S^2. This just the product of the projection of the tangent circle bundle onto S^2.

- But for SO(4) I am missing something. SO(4) acts transitively on S$$^{3}$$. The stabilizer of the north pole is a subgroup that preserves the equatorial 2-sphere and so is isomorphic to SO(3). Where is the mistake? This says that SO(4)/SO(3) is S$$^{3}$$.

This next argument leads to the same place. The differential of the action of SO(4) is transitive on the unit 2- sphere bundle of S$$^{3}$$. The differential of this action maps the unit circle bundle to the tangent 2-sphere bundle into itself transitively and with trivial stabilizers so SO(4) should be isomorphic to it. But this is S$$^{3}$$X SO(3) since the tangent circle bundle of the 2-sphere is diffeomorphic to SO(3) and the tangent bundle to S$$^{3}$$ is trivial.

 

[lavinia]

It depends on a holomorphic embedding, which we certainly don't have.
.

Is this why?
If the complex coordinate projections were holomorphic - as they are under the complex structure of CxC - then they must be constant on a holomorphic image of the compact surface . So this can not be an embedding. Yes?

 

[zhentil]

Is this why?
If the complex coordinate projections were holomorphic - as they are under the complex structure of CxC - then they must be constant on a holomorphic image of the compact surface . So this can not be an embedding. Yes?

Correct.

 

[zhentil]

In the first part you are saying that with a holomorphic embedding of the surface into R$$^{4}$$ one can pull back the tangent bundle of R$$^{4}$$ with its complex structure to get a trivial complex vector bundle over the surface. The tangent bundle to the surface is a complex subbundle and it's complement using a Hermitian metric - I suppose the metric is induced from R$$^{4}$$ - is another complex subbundle. Their Whitney sum is a trivial complex vector bundle.

- If the Chern class argument is wrong what can not happen? There is no holomorphic embedding of a surface into R$$^{4}$$ no matter what complex structure the surface and R$$^{4}$$ are given?

In the second part, I see that (SO(3) x SO(3)) / (U(1) x U(1)) is diffeomorphic to S^2 x S^2. This just the product of the projection of the tangent circle bundle onto S^2.

- But for SO(4) I am missing something. SO(4) acts transitively on S$$^{3}$$. The stabilizer of the north pole is a subgroup that preserves the equatorial 2-sphere and so is isomorphic to SO(3). Where is the mistake? This says that SO(4)/SO(3) is S$$^{3}$$.

This next argument leads to the same place. The differential of the action of SO(4) is transitive on the unit 2- sphere bundle of S$$^{3}$$. The differential of this action maps the unit circle bundle to the tangent 2-sphere bundle into itself transitively and with trivial stabilizers so SO(4) should be isomorphic to it. But this is S$$^{3}$$X SO(3) since the tangent circle bundle of the 2-sphere is diffeomorphic to SO(3) and the tangent bundle to S$$^{3}$$ is trivial.

SO(4)/SO(3) is equal to S^3. It's also equal to SO(3). Implicit in this is that we're given an action - the action (and hence the quotient) are different.

For the second part, you have to be more explicit about your action. There are many copies of SO(3) inside SO(4), and it's not clear to me that the specific stabilizer group in question acts transitively on the two-sphere bundle portion of S^3 x SO(3).

 

[lavinia]

For the second part, you have to be more explicit about your action. There are many copies of SO(3) inside SO(4), and it's not clear to me that the specific stabilizer group in question acts transitively on the two-sphere bundle portion of S^3 x SO(3).

OK. BTW: I appreciiate your patient help.

What I was thinking is to generalize what happens with SO(3). SO(3) acts on the 2 sphere by rotation and the defferential of its action acts on the unit circle bundle. This action is transitive and has trivial stabilizers. Thus SO(3) is diffeomorphic to the tangent circle bundle of the 2 sphere. (This gives another proof that the 2-sphere has no non-zero vector field since the real homology of SO(3) is zero in dimension 2.)

For SO(4) the differential of its action on the 3-sphere is an action on the unit tangent 2- sphere bundle and the stabilizer of the north pole acts by rotation on the tangent 2-sphere.

The differential of this differential action preserves the tangent unit circles to the tangent 2-spheres. This action has trivial stabilizers and is transitive ... I think.